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I have seen a question as;

If $\overline{AB}$ is a two-digit number

and

$\overline{AB}-10B-A=A^3 - B^3$

How many different AB two-digit numbers can be written?

Thank you.

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    Are you thinking of "$AB$" as the concatenation of two digits, or as the product of two numbers? (The edit suggests the former.) – 2017-01-20
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    I am also not sure about that point. The question is exactly like i expressed. The solution says that 10 different numbers can be written however i just found only "30" when i supposed AB is two-digit number. – 2017-01-20
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    If "$AB$" means $10A+B$ with $1\le A\le 9$ and $0\le B\le 9$, then you simply have $9(A-B)=A^3-B^3$. – 2017-01-20

2 Answers 2

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$\overline{AB}-10B-A=A^3-B^3\\(10A+B)-10B-A=A^3-B^3\\9(A-B)=(A-B)(A^2+B^2+AB)\\SO\\$ $$\begin{cases}A=B & \rightarrow A=B=1,2,3,4,...9\\A^2+B^2+AB=9 & \begin{cases}A=1 & B^2+1B=9-1\\A=2 & B^2+2B=9-4 \rightarrow \\A=3 & B^2+3B=9-9 \rightarrow B=0\end{cases} \end{cases} \\\overline{AB}=11,22,33,...,99,30$$

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    Brilliant. Thank you. – 2017-01-20
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$$ \begin{align} &10A+B-10B-A=A^3-B^3\\ \Rightarrow\ &\frac{(A^2+B^2+AB)(A-B)}{A-B}=9\\ \Rightarrow\ &A^2+B^2+AB=9\ or\ A=B \\ \end{align} $$

The solution if $A\ne B$ is $\overline{AB}=30$ i.e. $A=3,B=0$.

For $A=B$, there are $9$ solutions i.e. $11,22,...,99$.

Hence there are $10$ solutions.